# What would keep Dyson Swarm reflectors in place?

I've been reading a bit about the concept of a Dyson Swarm and how one of the more plausible routes would be to disassemble Mercury to construct ~1km hematite mirrors as the swarm reflectors. The basic concept is that the mirrors would be un-powered and simply reflect sunlight to a central processor that would extract the energy and beam it wherever we need. But that got me wondering; in such a scenario, wouldn't that simply turn the mirror elements into solar sails? Wouldn't they need an equivalent amount of energy in the opposite direction simply to keep them in place?

• So you don't have a link to where you read that version, right? Can you mention an author or a book it was in? – kim holder Aug 26 '16 at 2:33
• @kimholder I've added a couple, sorry for the omission. – thanby Aug 29 '16 at 14:47

If each 'mirror satellite' in the swarm could at least keep the same face towards the sun, then it wouldn't need to expend energy to stay in its orbit. This is because of a handy cancellation.

The force of gravity scales as $-m_{sat}/r^2$ and the force of the photon pressure scales as $+A_{sat}/r^2$ where $m_{sat}$ is the mass of the satellite (which might change a little if it has propellant for attitude control) and $A_{sat}$ is the effective area of the satellite. Since they both vary as $1/r^2$ the photon pressure just feels like the gravity of the sun is a little weaker.

Here are the full equations. For Radiation Pressure of a flat reflector with 100% reflectivity (no absorption/re-radiation):

$$P_{rad} = \frac{2E_f}{c} cos^2\alpha$$

where $E_f$ is the incident energy flux density of the sun's photons on the satellite (and $E_f/c$ is the incident momentum density thereof) and $\alpha$ is the tilt angle away from 'straight back towards the sun'. The factor of 2 in momentum comes from the reflection - you reverse the sign of the momentum $p$ of the photon so the mirror gets $2p$.

The force is then just pressure times area:

$$F_{rad} = A_{sat} P_{rad}$$

The energy flux density at the satellite's distance $r$ is just the total radiation power of the sun $\mathscr {P}_{sun}$ divided by the area of a sphere at that distance:

$$E_f = \mathscr {P}_{sun} \frac{1}{4 \pi r^2}$$

All together, assuming normal incidence (setting $cos^2 \alpha = 1)$:

$$F_{rad} = \frac { \mathscr {P}_{sun}}{2 \pi c} \frac{A_{sat}}{r^2}$$

The gravitational force comes from the 'modern form' of Newton's law of gravitation:

$$F_{grav} = -Gm_{sun}\frac{m_{sat}}{r^2}$$

So the same $1/r^2$ dependencies are in both forces, and since the signs are opposite, it simply "feels" like slightly weaker gravity.

side note: the cancellation sort-of works for elliptical orbits too, except for the libration. As the mirror orbits (revolves) around the Sun once, it also rotates around its axis once so that it points towards the sun all the time. But if it's in an elliptical orbit, it is sometimes moving faster and sometimes slower, but the rotation of the mirror around it's axis is constant. For very small eccentricities this 'periodic pointing error' will be 'very very' small (since $cos^2 \alpha$ is quartic about $\alpha = 0$ ), but it's not zero.

However if you can somehow keep the mirror pointed directly back at the sun, then the cancellation works just fine.

A graphic example of libration - the moon is rotating about its axis pretty much uniformly, but because of its elliptical orbit around the earth, it appears to be tilting back and forth. When it's farther away, it appears to be turning to the right faster, when it's close, it can't keep up and appears to move to the left.

Lunar Libration image from here At a given circular orbital radius, a shiny flat mirror pointing back towards the sun has a slightly lower orbital velocity that a diffuse white thing that scatters in many direction, and that has a slightly lower velocity than a dark thing that soaks up the power then re-radiates it isotropically. But they all just continue in their stable orbits (unless of course they rotate then it's a total mess!)

But the problem is that the satellites may change reflectivity due to damage, or effective area and reflection direction due to attitude changes, and the mass may change, and of course they all have gravitational effects on each other.

The ring can accommodate some of those, if they all behave identically. But the more complicated multi-ring orbits will perturb each other both gravitationally and by blocking each other's photon pressure, and that may require some much fancier orbits.

• Thanks for the detailed explanation! So given ideal conditions, it shouldn't be too hard (relatively speaking) to get the forces to cancel out for an orbiting ring of reflectors? – thanby Aug 29 '16 at 14:57
• @thanby well for anyone who can "disassemble Mercury", keeping the radiation pressure constant or at least only slowly decreasing should be fine. As the mirrors degrade, their reflectivity will decrease. but if it's over a long time, they would just very slowly spiral out a tiny bit, but remain pretty much circular. There might be issues with instabilities due to gravity between the mirrors in the right, but that should be a separate question. – uhoh Aug 29 '16 at 15:03

In the Wikipedia article you referenced, the Dyson Swarm constructs orbit around the star - standard orbital mechanics would apply, but traffic management might be interesting.

Further down in the same article, the Dyson Bubble has non-orbiting (& extremely lightweight) constructs that balance the solar sail pressure with gravity.

• How in the world... So to "deploy" this swarm, you'd have to orbit around the sun, stop completely, deploy the sail, then move on. You'd need some crazy propulsion systems to pull that off. I wonder what the size of the solar sail would need to be as well to keep it completely stationary (if we'd assume a couple tons for each mirror). – Magic Octopus Urn Jun 11 at 14:33